0 votes 0 votes find the probability that when four numbers from 1 to 100,inclusive,are picked at random with no repetitions allowed,either all are odd,all divisible by 3,or all divisible by 5 Mathematical Logic discrete-mathematics combinatory + – rahul sharma 5 asked Jun 11, 2017 retagged Jun 27, 2017 by Arjun rahul sharma 5 758 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply rahul sharma 5 commented Jun 11, 2017 reply Follow Share i got C(73/,4)/C(100,4). I it correct? 1 votes 1 votes Nachiket Karambelkar commented Jun 20, 2017 reply Follow Share yes i also got the same.. 0 votes 0 votes air1 commented Jun 20, 2017 i edited by air1 Jun 20, 2017 reply Follow Share By inclusion-exclusion principle I got, $\binom{50}{4} + \binom{33}{4} + \binom{20}{4} - \binom{17}{4} - \binom{10}{4} - \binom{6}{4} + \binom{3}{4}$. But I think, $\binom{a}{r} + \binom{b}{r} \ne \binom{a+b}{r}$ 1 votes 1 votes Please log in or register to add a comment.